Finding line of intersection of given plane and xz-plane. Please help

[saadatsubs]

Homework Statement

Given the plane: 4x - 2y +8z = 8,
Find Parametric equations of the line of intersection of the given plane and the xz-plane.

The answer for the question is: x = 2 +2t ; y = 0, z = -t

I don't understand how to get to this answer and I am going to be tested on this type of question. Can anyone help me understand this please?

Relevant Equations

x = x1 + at
y = y1 + bt
z = z1 +ct

Sorry for the really messy work I know I have a problem.

The other questions that the problem asked before the one I need help with are as follows:
Find the intercepts and sketch the plane.
Find the distance between the plane and the point (1,2,3)
Find the angle between the plane and the xz plane

Thank you for any help you can give me

 

[BvU]

Hello saadatsubs, $\qquad$ $\qquad$ !

I don't understand how to get to this answer

Aren't you thinking this through too deeply ?
Is it clear that points on the intersection with the xz-plane have y = 0 ? From there $$ 4x - 2y +8z = 8 \ \& \ y = 0 \Rightarrow 4x + 8z = 8 \Rightarrow x + 2z = 2 $$ and any choice for a parameter will do. You say 'the answer', but it's just an answer out of many. Other examples: x = t, y = 0 , z = (t-2)/2, or
z = t , y = 0, x = 2 - 2t etc, etc.

Can't read your picture: stiff neck.

 

[LCKurtz]

A general method to find the line of intersection of two planes is:
1. The cross product of the two normals to the planes gives a direction vector for the line.
2. Find any point on both planes to use with the direction vector. You can usually set one variable = 0 and solve the remaining two equations for the other coordinates of the point.